Question: $h(x)=\begin{cases} x^2-1&\text{for }x\leq3 \\\\ 2x+1&\text{for }3<x<10 \end{cases}$ Find $\lim_{x\to 3^-}h(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $3$ (Choice B) B $7$ (Choice C) C $8$ (Choice D) D The limit doesn't exist.
Answer: Notice that we were asked to find the one-sided limit, $\lim_{x\to 3^-}h(x)$. This is the limit where $x$ -values approach $3$ from the left. Let's find the limit as $x$ approaches $3$ from the left. We will use the fact that $h(x)=x^2-1$ for $x$ -values smaller than $3$. $\begin{aligned} &\phantom{=}\lim_{x\to 3^-}h(x) \\\\ &=\lim_{x\to 3^-}x^2-1 \\\\ &=(3)^2-1&\gray{\text{Direct substitution}} \\\\ &=8 \end{aligned}$ In conclusion, we found that $\lim_{x\to 3^-}h(x)=8$.